# A triangle has sides with lengths: 2, 9, 2. How do you find the area of the triangle using Heron's formula?

Dec 26, 2015

There is no such triangle, since $2 + 2 < 9$

#### Explanation:

If a triangle has sides of length $a$, $b$ and $c$ then all of these conditions hold:

$a + b > c$

$b + c > a$

$c + a > b$

...unless you count empty triangles, in which case change the $>$'s into $\ge$'s.

If you try to apply Heron's formula to lengths $a = 2$, $b = 9$, $c = 2$, then you will find that you end up attempting to take the square root of a negative number, hence no Real area:

The semi-perimeter $s p$ is given by:

$s p = \frac{a + b + c}{2} = \frac{2 + 9 + 2}{2} = \frac{13}{2}$

Then Heron's formula for the area $A$ is:

$A = \sqrt{s p \left(s p - a\right) \left(s p - b\right) \left(s p - c\right)}$

$= \sqrt{\frac{13}{2} \left(\frac{13}{2} - 2\right) \left(\frac{13}{2} - 9\right) \left(\frac{13}{2} - 2\right)}$

$= \sqrt{\left(\frac{13}{2}\right) \left(\frac{9}{2}\right) \left(- \frac{5}{2}\right) \left(\frac{9}{2}\right)}$

$= \sqrt{- \frac{5265}{16}}$

It is possible to simplify this further, but there's no real point since it's clearly the square root of a negative quantity.